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Drupady [299]
3 years ago
9

Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

ght travels in one year. State the speed in mi/s, using scientific notation. (1 mile = 5280 feet exactly, by definition; i.e., this is not a measurement.) A. 1.8600 x 105 mi/s Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance light travels in one year.
State the speed in mi/s, using scientific notation. (1 mile = 5280 feet exactly, by definition; i.e., this is not a measurement.)

A. 1.8600 x 105 mi/s
B. 5.1854 x 1012 mi/s
Physics
1 answer:
larisa86 [58]3 years ago
6 0

A. 1.8600 x 105 mi/s

Explanation:

The problem gives us the value of the speed of light in ft/s: v=982,080,000 ft/s

We know the conversion factor between feet and miles:

1 mi = 5280 ft

So we can apply this conversion factor to the speed of light:

v=982,080,000 ft/s \cdot \frac{1}{5280 ft/mi}=186,000 mi/s=1.86\cdot 10^5 mi/s


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1 year ago
A man starts walking north at 3 ft/s from a point P. Five minutes later a woman starts walking south at 4 ft/s from a point 500
mrs_skeptik [129]

Answer:

ds/dt = 6.98 ft/s

Explanation:

Given:

- The speed of man due north Vm = 3 ft/s

- The speed of woman due south Vw = 4 ft/s

- Woman starts walking 5 mins later than man

Find:

At what rate are the people moving apart 15 min after the woman starts walking?

Solution:

- The total time for which the man is walking due north from P, is Tm:

                                   Tm = 5 + 15 = 20 mins

- The total distance traveled by man in Tm mins is:

                                   Dm = Tm*Vm

                                   Dm = 20*60*3

                                   Dm = 3,600 ft

- The total time for which the woman is walking due south from 500 ft due east from P, is Tw:

                                   Tw = 15 = 15 mins

- The total distance traveled by man in Tw mins is:

                                   Dw = Tw*Vw

                                   Dw = 15*60*4

                                   Dw = 3,600 ft

- The displacement between man and woman at any instance is (s) which can be related by pythagoras theorem as follows:

                                   s^2 = (dm + dw)^2 + 500^2

Where, dm : Distance travelled by man at any time Tm

            dw : Distance travelled by woman at any time Tw

- Differentiate s with respect to t:

                                   2s*ds/dt = 2*(dm + dw)*(Vm + Vw)

                                   s*ds/dt = (dm + dw)*(Vm + Vw)

                                   ds/dt = [ (dm + dw)*(Vm + Vw) ] / s

- Evaluate the rate of separation of man and woman ds/dt by evaluating at instance Tm = 20 mins and Tw = 15 mins. We have:

                 ds/dt = [ (Dm + Dw)*(Vm + Vw) ] / sqrt ( (Dm + Dw)^2 + 500^2 )

- Plug in the values:

                 ds/dt = [ (3600 + 3600)*(3 + 4) ] / [sqrt ( (3600 + 3600)^2 + 500^2 )]  

                ds/dt = 6.98 ft/s

                 

           

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